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3 years ago

In terms of newton first law of motion why is it important to wear a seat belt

Physics

1 answer:

lilavasa [31]3 years ago

6 0

Answer:

Seatbelts stop you

Explanation:

Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

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How many significant figures are in the measurement 0.020 km?

Volgvan

Answer:3

Explanation:

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2 years ago

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A small 24 kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?

USPshnik [31]

J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј

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3 years ago

What is power, and what is its relationship to voltage and amperage? (4 points)

tigry1 [53]

Answer:

The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

Explanation: brainliest pls

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3 years ago

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Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat

aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0

2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

3 years ago